Maximization problem — Stiffest beam that can be cut from a log

[Karol]

I made a small mistake, again:
$$\eta=\frac{1}{3},~a^2=\eta b^2~\rightarrow~b^2=3a^2,~a^2+b^2=r^2~\rightarrow~a^2+3a^2=r^2~\rightarrow~a=\frac{r}{2}$$
$a~$ should equal $~r$

 

[StoneTemplePython]

I made a small mistake, again:
$$\eta=\frac{1}{3},~a^2=\eta b^2~\rightarrow~b^2=3a^2,~a^2+b^2=r^2~\rightarrow~a^2+3a^2=r^2~\rightarrow~a=\frac{r}{2}$$
$a~$ should equal $~r$

I you want to be very literal about this, consider that, after re-examining the picture, what we actually have as our constraint is that

$a^2 + b^2 = d^2$ where $d$ is the diameter -- I was being loose by using $r$ (though it works up to a rescaling) but, again, the literal interpretation is to use $d$ as the diameter for the constraint. None of the inequality work changes after substituting in $d$ in place of $r$. Using your work, but with the substituted value, you get:

$$\eta=\frac{1}{3},~a^2=\eta b^2~\rightarrow~b^2=3a^2,~a^2+b^2=d^2~\rightarrow~a^2+3a^2=d^2~\rightarrow~a=\frac{d}{2} = r$$

 

[epenguin]

I drew a=r:
View attachment 221572

I found $\eta=\frac{1}{3}~$, not 3.
The right side isn't important. it's not important what the stiffness is as long as it's the maximum.
$$\eta=\frac{1}{3},~a^2=\eta b^2~\rightarrow~b^2=3a^2,~a^2+b^2=r^2~\rightarrow~a^2+3b^2=r^2~\rightarrow~a=\frac{r}{2}$$
Wrong. in this answer a is the whole width

OK The whole with can equal the radius.And the answer to this problem is that it does. The radius and the base form an equilateral triangle. Trouble is you seem seem never to have made up your mind whether a and b represent sides or half sides of the rectangle. In this very post you seem to use both meanings. I don't know what this whole thread is about, except confusion caused by this, certainly all these γ's and η's are unnecessary. Let $x, y$ mean half side, then
$$S=16kxy^3$$
I will just find the $x$ that gives a maximum of $xy^3$ which I call
$$s=xy^3\\s=x\left( r^{2}-x^{2}\right) ^{3/2}\\
s^\prime=\left( r^{2}-x^{2}\right) ^{3/2}-3x^{2}\left( r^{2}-x^{2}\right) ^{1/2}\\
=\left( r^{2}-x^{2}\right) ^{1/2}\left( r^{2}-4x^{2}\right)
$$

So the maximum is where $x=r/2$, ...

I get S_max = 3√3$~r^4$

 

[Karol]

I thank you:
Math_QED
phinds
Orodruin
StoneTemplePhyton
epenguin
Ray Vickson